For all z inside of C (C the unit circle oriented counterclockwise),
[tex]
f(z) = \frac{1}{2\pi i}\int_C \frac{g(u)}{u-z} du
[/tex]
where [itex]g(u) = \bar{u}[/itex] is a continuous function and [itex]f[/itex] is analytic in C. Describe [itex]f[/itex]in C in terms of a power series.

[itex]\displaystyle f(z) = \frac{1}{2\pi i}\int_C \frac{\bar{u}}{u-z} du[/itex] I am confused with what I am supposed to do. I know it says describe [itex]f[/itex] in terms of a power series.

All of the terms in your series are zero, yes. But that still doesn't make f(z)=1/z=0. I'm not sure you are paying attention here.

I understand what you are saying but I am trying to solve for
$$
f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}\left[\int_0^{2\pi}\frac{\frac{1}{u}}{u^{n + 1}}du z^n\right].
$$
Since Laurent series are out and all the terms are 0, what else could f(z) be?

I understand what you are saying but I am trying to solve for
$$
f(z) = \frac{1}{2\pi i}\sum_{n = 0}^{\infty}\left[\int_0^{2\pi}\frac{\frac{1}{u}}{u^{n + 1}}du z^n\right].
$$
Since Laurent series are out and all the terms are 0, what else could f(z) be?

I am trying to tell you that the series you are quoting is NOT valid for all functions f(z). f(z) has to be analytic at z=0 to apply that. f(z)=1/z is NOT analytic at z=0. I've already told you this.

By the integral transform theorem, if you put a continuous function g(u) into f(z), you get out an analytic function. If g is analytic, you get the same function. So f(z) has to equal something.

I'm not quite sure why this is so difficult. 1/u is continuous on the contour. And yes, you get an analytic function out. It's f(z)=0. Now you say "If g is analytic, you get the same function.". g(u)=1/u ISN'T analytic at u=0. So the function you get out f(z)=0, ISN'T the same as the function you put in f(z)=1/z.